Selection of Terms in an Arithmetic Progression

Sometimes we need to assume certain number of terms in Arithmetic Progression. The following ways are generally used for the selection of terms in an arithmetic progression.

(i) If the sum of three terms in Arithmetic Progression be given, assume the numbers as a - d, a and a + d. Here common difference is d.

(ii) If the sum of four terms in Arithmetic Progression be given, assume the numbers as a - 3d, a - d, a + d and a + 3d.

(iii) If the sum of five terms in Arithmetic Progression be given, assume the numbers as a - 2d, a - d, a, a + d and a + 2d. Here common difference is 2d.

(iv) If the sum of six terms in Arithmetic Progression be given, assume the numbers as a - 5d, a - 3d, a - d, a + d, a + 3d and a + 5d. Here common difference is 2d.

Note: From the above explanation we understand that in case of an odd number of terms, the middle term is ‘a’ and the common difference is ‘d’.

Again, in case of an even number of terms the middle terms are a - d, a + d and the common difference is 2d.


Solved examples to observe how to use the selection of terms in an arithmetic progression

1. The sum of three numbers in Arithmetic Progression is 12 and the sum of their square is 56. Find the numbers.

Solution:

Let us assume that the three numbers in Arithmetic Progression be a - d, a and a + d.

According to the problem,

Sum = 12                             and

⇒ a - d + a + a + d = 12

⇒ 3a = 12

⇒ a = 4

Sum of the squares = 56

(a - d)\(^{2}\) + a\(^{2}\) + (a + d)\(^{2}\) = 56

⇒ a\(^{2}\) - 2ad + d\(^{2}\) + a\(^{2}\) + a\(^{2}\) + 2ad + d\(^{2}\) = 56

⇒ 3a\(^{2}\) + 2d\(^{2}\) = 56

⇒ 3 × (4)\(^{2}\) + 2d\(^{2}\) = 56

⇒ 3 × 16 + 2d\(^{2}\) = 56

⇒ 48 + 2d\(^{2}\) = 56

⇒ 2d\(^{2}\) = 56 - 48

⇒ 2d\(^{2}\) = 8

⇒ d\(^{2}\) = 4

⇒ d = ± 2

If d = 3, the numbers are 4 – 2, 4, 4 + 2 i.e., 2, 4, 6

If d = -3, the numbers are 4 + 2, 4, 4 - 2 i.e., 6, 4, 2

Therefore, the required numbers are 2, 4, 6 or 6, 4, 2.

2. The sum of four numbers in Arithmetic Progression is 20 and the sum of their square is 120. Find the numbers.

Solution: 

Let us assume that the four numbers in Arithmetic Progression be a - 3d, a - d, a + d and a + 3d.

According to the problem,

Sum = 20

⇒ a - 3d + a - d + a + d + a + 3d = 20

⇒ 4a = 20

⇒ a = 5

and

Sum of the squares = 120

⇒ (a - 3d)\(^{2}\) + (a - d)\(^{2}\) + (a + d)\(^{2}\) + (a + 3d)\(^{2}\) = 120

⇒ a\(^{2}\) - 6ad + 9d\(^{2}\) + a\(^{2}\) - 2ad + d\(^{2}\) + a\(^{2}\) + 2ad + d\(^{2}\) + a\(^{2}\) + 6ad + 9d\(^{2}\) = 120

⇒ 4a\(^{2}\) + 20d\(^{2}\) = 120

⇒ 4 × (5)\(^{2}\) + 20d\(^{2}\) = 120

⇒ 4 × 25 + 20d\(^{2}\) = 120

⇒ 100 + 20d\(^{2}\) = 120

⇒ 20d\(^{2}\) = 120 - 100

20d\(^{2}\) = 20

⇒ d\(^{2}\) = 1

⇒ d = ± 1

If d = 1, the numbers are 5 - 3, 5 - 1, 5 + 1, 5 + 3 i.e., 2, 4, 6, 8

If d = -1, the numbers are 5 + 3, 5 + 1, 5 - 1, 5 - 3 i.e., 8, 6, 4, 2

Therefore, the required numbers are 2, 4, 6, 8 or 8, 6, 4, 2.

 

3. The sum of three numbers in Arithmetic Progression is -3 and their product is 8. Find the numbers.

Solution:

Let us assume that the three numbers in Arithmetic Progression be a - d, a and a + d.

According to the problem,

Sum = -3                                  and

⇒ a - d + a + a + d = -3

⇒ 3a = -3

⇒ a = -1

Product = 8

⇒ (a - d) (a) (a + d) = 8

⇒ (-1)[(-1)\(^{2}\) - d\(^{2}\)] = 8

⇒ -1(1 - d\(^{2}\)) = 8

⇒ -1 + d\(^{2}\) = 8

⇒ d\(^{2}\) = 8 + 1

⇒ d\(^{2}\) = 9

⇒ d = ± 3

If d = 3, the numbers are -1 - 3, -1, -1 + 3 i.e., -4, -1, 2

If d = -3, the numbers are -1 + 3, -1, -1 - 3 i.e., 2, -1, -4

Therefore, the required numbers are -4, -1, 2 or 2, -1, -4.

Arithmetic Progression





11 and 12 Grade Math 

From Selection of Terms in an Arithmetic Progression to HOME PAGE




Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.



New! Comments

Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.




Share this page: What’s this?

Recent Articles

  1. Addition of Decimals | How to Add Decimals? | Adding Decimals|Addition

    Apr 24, 25 01:45 AM

    Addition of Decimals
    We will discuss here about the addition of decimals. Decimals are added in the same way as we add ordinary numbers. We arrange the digits in columns and then add as required. Let us consider some

    Read More

  2. Addition of Like Fractions | Examples | Videos | Worksheet | Fractions

    Apr 23, 25 09:23 AM

    Adding Like Fractions
    To add two or more like fractions we simplify add their numerators. The denominator remains same. Thus, to add the fractions with the same denominator, we simply add their numerators and write the com…

    Read More

  3. Subtraction | How to Subtract 2-digit, 3-digit, 4-digit Numbers?|Steps

    Apr 23, 25 12:41 AM

    Subtraction Example
    The answer of a subtraction sum is called DIFFERENCE. How to subtract 2-digit numbers? Steps are shown to subtract 2-digit numbers.

    Read More

  4. Subtraction of 4-Digit Numbers | Subtract Numbers with Four Digit

    Apr 23, 25 12:38 AM

    Properties of Subtraction of 4-Digit Numbers
    We will learn about the subtraction of 4-digit numbers (without borrowing and with borrowing). We know when one number is subtracted from another number the result obtained is called the difference.

    Read More

  5. Subtraction with Regrouping | 4-Digit, 5-Digit and 6-Digit Subtraction

    Apr 23, 25 12:34 AM

     Subtraction of 5-Digit Numbers with Regrouping
    We will learn subtraction 4-digit, 5-digit and 6-digit numbers with regrouping. Subtraction of 4-digit numbers can be done in the same way as we do subtraction of smaller numbers. We first arrange the…

    Read More